by clicking the arrows at the side of the page, or by using the toolbar.
by clicking anywhere on the page.
by dragging the page around when zoomed in.
by clicking anywhere on the page when zoomed in.
web sites or send emails by clicking on hyperlinks.
Email this page to a friend
Search this issue
Index - jump to page or section
Archive - view past issues
Concrete In Australia : June 2013
Concrete in Australia Vol 39 No 2 19 vertical tensile stress, will be proportional to the proportion of load at the bottom so that if, for example wT = wB = 0.5*wu then, in the case of Figure 1: Part rfy = 12wu bw = 156 2*350 = 0.223 MPa e rest of this paper will be simplified by analysing Figure 1 forthe worstcase:wB = wu andwT=0. 7.0 STATIC EQUILIBRIUM OF THE FIRST FRAGMENT AND THE FIRST HINGE Figure 3 shows the bottom rebar failing, in anchorage, at the face of the support with an assumed anchorage strength T1 = 283 kN. e depth of the bending-compression stress-block is calculated as usual, notwithstanding the dogleg yield-line. e centroid of the bending-compression force is 19 mm below the top surface so the moment strength at hinge 1 is: M1 = 283 kN * 0.541 = 153 kNm. is moment strength must be balanced by the moment of the vertical forces so the lever-arm of those vertical forces must be: h10 = M1 R = 153*1000 391 = 391 mm. Lever-arm h10 is actually the average of the length of the bearing h0 and the horizontal projected length of the first hinge h1 so: h10 =h1+h0 2 and the horizontal projected length of the first hinge is: h1 = 2*h10 -- h0 = 2*391-300 = 482 mm. ere is no direct load on this fragment because all of the load is now assumed applied at the bottom. So V1 = R, requiring an ideal smeared vertical yield-strength: rfy1 = V1 bw *h1 = 391*1000 350*482 = 2.32 MPa for the vertical shear reinforcement across hinge 1. Assume N10 stirrups with three vertical legs per set (one outer stirrup plus one single leg tie on central N24) giving a tie- set area of Asv1 = 3*78.5 = 236 mm2 spaced at s1= Asv1 * fsy *(φV = 0.70) bw * rfy1 = 236*500*0.70 350*2.32 = 102 mm. 8.0 USE N10-100 (THREE VERTICAL LEGS) Figure 4 shows a lower-bound solution for the first segment. Note that the parabolic half-arch should have a horizontal tangent at the top right-hand end, notwithstanding this writer s indifferent CAD skills. e rules for drawing parabolic half-arches are the same/similar to those for drawing parabolic vertical curves for roads. 9.0 SECOND HANGER SEGMENT = FIRST TOOTH Figure 5 shows the second hanger segment calculated on the Figure 4. Lower-bound solution for first hanger segment. Figure 3. First hinge. Figure 5. Second hinge and first tooth.